Question: A secant line intersects the curve $y=-2x^2-7x$ at two points with $x$ -coordinates $-4$ and $t$, where $t\neq -4$. What is the slope of the secant line in terms of $t$ ? Your answer must be fully expanded and simplified.
Explanation: We are given that the secant line intersects the curve at $x=-4$ and $x=t$. Since these points are on the curve $y=-2x^2-7x$, we know that their $y$ -values are $y=-4$ and $y=-2t^2-7t$, correspondingly. To summarize this part, we know that the secant line passes through the points $(-4,-4)$ and $(t, -2t^2-7t)$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{-2t^2-7t-(-4)}{t-(-4)} \\\\ &=\dfrac{-2t^2-7t+4}{t+4} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} \dfrac{-2t^2-7t+4}{t+4}&=\dfrac{-1(2t^2+7t-4)}{t+4} \\\\ &=\dfrac{-1(2t-1)(t+4)}{t+4} \\\\ &=-1(2t-1)\text{, for }t\neq -4 \end{aligned}$ Since we are given that $t\neq -4$, we can conclude that the slope of the secant line is $-1(2t-1)$ or $1-2t$.